§7.prop.35ἐὰν δύο ἀριθμοὶ ἀριθμόν τινα μετρῶσιν, καὶ ὁ ἐλάχιστος ὑπʼ αὐτῶν μετρούμενος τὸν αὐτὸν μετρήσει. δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν ΓΔ μετρείτωσαν, ἐλάχιστον δὲ τὸν Ε· λέγω, ὅτι καὶ ὁ Ε τὸν ΓΔ μετρεῖ.
If two numbers measure some number, the least number measured by them will also measure the same. For let two numbers A, B measure some number ΓΔ, and let E be the least; I say that E also measures ΓΔ. For let two numbers A, B measure some number ΓΔ, and let E be the least; I say that E also measures ΓΔ.
εἰ γὰρ οὐ μετρεῖ ὁ Ε τὸν ΓΔ, ὁ Ε τὸν ΔΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΓΖ. καὶ ἐπεὶ οἱ Α, Β τὸν Ε μετροῦσιν, ὁ δὲ Ε τὸν ΔΖ μετρεῖ, καὶ οἱ Α, Β ἄρα τὸν ΔΖ μετρήσουσιν. μετροῦσι δὲ καὶ ὅλον τὸν ΓΔ· καὶ λοιπὸν ἄρα τὸν ΓΖ μετρήσουσιν ἐλάσσονα ὄντα τοῦ Ε· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οὐ μετρεῖ ὁ Ε τὸν ΓΔ· μετρεῖ ἄρα· ὅπερ ἔδει δεῖξαι.
For if E does not measure ΓΔ, let E by measuring ΔΖ leave ΓΖ less than itself. And since A, B measure E, and E measures ΔΖ, therefore A, B will also measure ΔΖ. But they also measure the whole ΓΔ; therefore they will also measure the remainder ΓΖ, which is less than E; which is impossible. Therefore it is not the case that E does not measure ΓΔ; therefore it measures it; which was to be proved. For if E does not measure ΓΔ, let E by measuring ΔΖ leave ΓΖ less than itself. And since A, B measure E, and E measures ΔΖ, therefore A, B will also measure ΔΖ. But they also measure the whole ΓΔ; therefore they will also measure the remainder ΓΖ, which is less than E; which is impossible. Therefore it is not the case that E does not measure ΓΔ; therefore it measures it; which was to be proved.
§7.prop.36τριῶν ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν. ἔστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ οἱ Α, Β, Γ· δεῖ δὴ εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν.
Given three numbers, to find the least number which they measure. Let the given three numbers be A, B, Γ; it is then required to find the least number which they measure. Let the given three numbers be A, B, Γ; it is then required to find the least number which they measure.
εἰλήφθω γὰρ ὑπὸ δύο τῶν Α, Β ἐλάχιστος μετρούμενος ὁ Δ. ὁ δὴ Γ τὸν Δ ἤτοι μετρεῖ ἢ οὐ μετρεῖ. μετρείτω πρότερον. μετροῦσι δὲ καὶ οἱ Α, Β τὸν Δ· οἱ Α, Β, Γ ἄρα τὸν Δ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσιν ἀριθμὸν οἱ Α, Β, Γ ἐλάσσονα ὄντα τοῦ Δ. μετρείτωσαν τὸν Ε. ἐπεὶ οἱ Α, Β, Γ τὸν Ε μετροῦσιν, καὶ οἱ Α, Β ἄρα τὸν Ε μετροῦσιν. καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Α, Β μετρούμενος μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Α, Β μετρούμενός ἐστιν ὁ Δ· ὁ Δ ἄρα τὸν Ε μετρήσει ὁ μείζων τὸν ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον.
For let Δ be taken as the least number measured by the two numbers A, B. Then Γ either measures Δ or does not measure it. First, let it measure it. But A, B also measure Δ; therefore A, B, Γ measure Δ. I say then that they also measure the least. For if not, A, B, Γ will measure a number which is less than Δ. Let them measure E. Since A, B, Γ measure E, therefore A, B also measure E. Therefore the least number measured by A, B will also measure it. But Δ is the least number measured by A, B; therefore Δ will measure E, the greater the less; which is impossible.
οὐκ ἄρα οἱ Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Δ· οἱ Α, Β, Γ ἄρα ἐλάχιστον τὸν Δ μετροῦσιν. μὴ μετρείτω δὴ πάλιν ὁ Γ τὸν Δ, καὶ εἰλήφθω ὑπὸ τῶν Γ, Δ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Ε. ἐπεὶ οἱ Α, Β τὸν Δ μετροῦσιν, ὁ δὲ Δ τὸν Ε μετρεῖ, καὶ οἱ Α, Β ἄρα τὸν Ε μετροῦσιν. μετρεῖ δὲ καὶ ὁ Γ οἱ Α, Β, Γ ἄρα τὸν Ε μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί τινα οἱ Α, Β, Γ ἐλάσσονα ὄντα τοῦ Ε. μετρείτωσαν τὸν Ζ. ἐπεὶ οἱ Α, Β, Γ τὸν Ζ μετροῦσιν, καὶ οἱ Α, Β ἄρα τὸν Ζ μετροῦσιν· καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Α, Β μετρούμενος τὸν Ζ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Α, Β μετρούμενός ἐστιν ὁ Δ· ὁ Δ ἄρα τὸν Ζ μετρεῖ. μετρεῖ δὲ καὶ ὁ Γ τὸν Ζ· οἱ Δ, Γ ἄρα τὸν Ζ μετροῦσιν· ὥστε καὶ ὁ ἐλάχιστος ὑπὸ τῶν Δ, Γ μετρούμενος τὸν Ζ μετρήσει. ὁ δὲ ἐλάχιστος ὑπὸ τῶν Γ, Δ μετρούμενός ἐστιν ὁ Ε· ὁ Ε ἄρα τὸν Ζ μετρεῖ ὁ μείζων τὸν ἐλάσσονα· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Ε. ὁ Ε ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β, Γ μετρεῖται· ὅπερ ἔδει δεῖξαι.
Therefore A, B, Γ will not measure any number which is less than Δ; therefore A, B, Γ measure Δ as the least. For let Δ be taken as the least number measured by the two numbers A, B. Then Γ either measures Δ or does not measure it. First, let it measure it. But A, B also measure Δ; therefore A, B, Γ measure Δ. I say then that they also measure the least. For if not, A, B, Γ will measure a number which is less than Δ. Let them measure E. Since A, B, Γ measure E, therefore A, B also measure E. Therefore the least number measured by A, B will also measure it. But Δ is the least number measured by A, B; therefore Δ will measure E, the greater the less; which is impossible. Therefore A, B, Γ will not measure any number which is less than Δ; therefore A, B, Γ measure Δ as the least. Again, let Γ not measure Δ, and let E be taken as the least number measured by Γ, Δ. Since A, B measure Δ, and Δ measures E, therefore A, B also measure E. But Γ also measures it; therefore A, B, Γ measure E. I say then that they also measure the least. For if not, A, B, Γ will measure some number which is less than E. Let them measure Z. Since A, B, Γ measure Z, therefore A, B also measure Z; therefore also the least number measured by A, B will measure Z. But Δ is the least number measured by A, B; therefore Δ measures Z. But Γ also measures Z; therefore Δ, Γ measure Z; so that the least number measured by Δ, Γ will also measure Z. But the least number measured by Γ, Δ is E; therefore E measures Z, the greater the less; which is impossible. Therefore A, B, Γ will not measure any number which is less than E. Therefore E is the least number measured by A, B, Γ; which was to be proved. Again, let Γ not measure Δ, and let E be taken as the least number measured by Γ, Δ. Since A, B measure Δ, and Δ measures E, therefore A, B also measure E. But Γ also measures it; therefore A, B, Γ measure E. I say then that they also measure the least. For if not, A, B, Γ will measure some number which is less than E. Let them measure Z. Since A, B, Γ measure Z, therefore A, B also measure Z; therefore also the least number measured by A, B will measure Z. But Δ is the least number measured by A, B; therefore Δ measures Z. But Γ also measures Z; therefore Δ, Γ measure Z; so that the least number measured by Δ, Γ will also measure Z. But the least number measured by Γ, Δ is E; therefore E measures Z, the greater the less; which is impossible. Therefore A, B, Γ will not measure any number which is less than E. Therefore E is the least number measured by A, B, Γ; which was to be proved.